KCET · Maths · Functions
Domain of the function \(f(x)=\frac{1}{\sqrt{\left[x^{2}\right]-[x]-6}}\)
where \([x]\) is greatest integer \(\leq x\) is
- A \((-\infty,-2) \cup[4, \infty)\)
- B \((-\infty,-2) \cup[3, \infty]\)
- C \([-\infty,-2] \cup[4, \infty]\)
- D \([-\infty,-2] \cup[3, \infty)\)
Answer & Solution
Correct Answer
(A) \((-\infty,-2) \cup[4, \infty)\)
Step-by-step Solution
Detailed explanation
Given function,
\(f(x)=\frac{1}{\sqrt{[x]^{2}-[x]-6}}\)
where \([x]\) is greatest integer \(\leq x\).
\(\begin{array}{ll} & {[x]^{2}-[x]-6>0} \\ \Rightarrow & {[x]^{2}-3[x]+2[x]-6>0} \\ \Rightarrow & ([x]-3)([x]+2)>0 \\ \Rightarrow & ([x]-3)>0,[x]+2 < 0 \\ \Rightarrow & {[x]>3,[x] < -2} \\ \Rightarrow & x \in[4, \infty), x \in(-\infty,-2) \\ \therefore & x \in(-\infty,-2) \cup[4, \infty)\end{array}\)
\(f(x)=\frac{1}{\sqrt{[x]^{2}-[x]-6}}\)
where \([x]\) is greatest integer \(\leq x\).
\(\begin{array}{ll} & {[x]^{2}-[x]-6>0} \\ \Rightarrow & {[x]^{2}-3[x]+2[x]-6>0} \\ \Rightarrow & ([x]-3)([x]+2)>0 \\ \Rightarrow & ([x]-3)>0,[x]+2 < 0 \\ \Rightarrow & {[x]>3,[x] < -2} \\ \Rightarrow & x \in[4, \infty), x \in(-\infty,-2) \\ \therefore & x \in(-\infty,-2) \cup[4, \infty)\end{array}\)
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